Electronic Devices (Test 2)

Gate Exam : Ec Electronics And Communication Engineering

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Electronic Devices

Electronic Devices
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Q.1
For the BJT circuit shown, assume that the β of the transistor is very large and VBE = . V. The mode of operation of the BJT is

A. cut-off
B. saturation
C. normal active
D. reverse active
Answer : Option B
Explaination / Solution:



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Q.2
Consider the following two statements about the internal conditions in a n - channel MOSFET operating in the active region.
S1 : The inversion charge decreases from source to drain
S2 : The channel potential increases from source to drain.
Which of the following is correct?
A. Only S2 is true
B. Both S1 and S2 are false
C. Both S1 and S2 are true, but S2 is not a reason for S1
D. Both S1 and S2 are true, and S2 is a reason for S1
Answer : Option D
Explaination / Solution:

Both S1 and S2 are true and S2 is a reason for S1.

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Q.3
Which of the following is NOT associated with a p - n junction ?
A. Junction Capacitance
B. Charge Storage Capacitance
C. Depletion Capacitance
D. Channel Length Modulations
Answer : Option D
Explaination / Solution:

Channel length modulation is not associated with a p - n junction. It is being associated with MOSFET in which effective channel length decreases, producing the phenomenon called channel length modulation.

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Q.4
A silicon wafer has 100 nm of oxide on it and is furnace at a temperature above 10000 C for further oxidation in dry oxygen. The oxidation rate
A. is independent of current oxide thickness and temperature
B. is independent of current oxide thickness but depends on temperature
C. slows down as the oxide grows
D. is zero as the existing oxide prevents further oxidation
Answer : Option D
Explaination / Solution:
No Explaination.


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Q.5
The measured trans conductance gm of an NMOS transistor operating in the linear region is plotted against the gate voltage VG at a constant drain voltage VD. Which of the following figures represents the expected dependence of gm on VG ?
A.
B.
C.
D.
Answer : Option C
Explaination / Solution:
No Explaination.


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Q.6
The cross section of a JFET is shown in the following figure. Let Vc be -2 V and let VP be the initial pinch -off voltage. If the width W is doubled (with other geometrical parameters and doping levels remaining the same), then the ratio between the mutual trans conductances of the initial and the modified JFET is

A. 4
B.
C.
D.
Answer : Option C
Explaination / Solution:



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Q.7
In a forward biased pn junction diode, the sequence of events that best describes the mechanism of current flow is
A. injection, and subsequent diffusion and recombination of minority carriers
B. injection, and subsequent drift and generation of minority carriers
C. extraction, and subsequent diffusion and generation of minority carriers
D. extraction, and subsequent drift and recombination of minority carriers
Answer : Option A
Explaination / Solution:

The potential barrier of the pn junction is lowered when a forward bias voltage is applied, allowing electrons and holes to flow across the space charge region (Injection) when holes flow from the p region across the space charge region into the n region, they become excess minority carrier holes and are subject to diffuse, drift and recombination processes. Hence correct option is (A)

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Q.8
In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency. vi is a small signal input. The gain magnitude  at 10 M rad/s is

A. maximum
B. minimum
C. unity
D. zero
Answer : Option A
Explaination / Solution:

For the parallel RLC circuit resonance frequency is,

Thus given frequency is resonance frequency and parallel RLC circuit has maximum impedance at resonance frequency
Gain of the amplifier is   where ZC is impedance of parallel RLC circuit.

Hence at this frequency (ωr), gain is
 which is maximum
Therefore gain is maximum at ωr = 10 /sec M rad .

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Q.9
In the silicon BJT circuit shown below, assume that the emitter area of transistor Q1 is half that of transistor Q2

The value of current I0 is approximately
A. 0.5 mA
B. 2 mA
C. 9.3 mA
D. 15 mA
Answer : Option B
Explaination / Solution:

Since, emitter area of transistor Q1 is half of transistor Q2, so current


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Q.10
For a BJT the common base current gain α = 0.98 and the collector base junction reverse bias saturation current ICO = 0.6 μA. This BJT is connected in the common emitter mode and operated in the active region with a base drive current IB = 204 A. The collector current IC for this mode of operation is
A. 0.98 mA
B. 0.99 mA
C. 1.0 mA
D. 1.01 mA
Answer : Option D
Explaination / Solution:


In active region, for common emitter amplifier,

Substituting ICO = 0.6 μA and IB = 204 μA in above eq we have, 
IC = 1.01 mA

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EC Electronics and Communication Engineering